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M408L_HW10 - hyun(hh7953 – HW10 – gogolev –(57440 1...

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Unformatted text preview: hyun (hh7953) – HW10 – gogolev – (57440) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay 1 integraldisplay 2 1 (2 x + 3 x 2 y ) dydx . 1. I = 5 2 correct 2. I = 3 2 3. I = 2 4. I = 7 2 5. I = 3 Explanation: The integral can be written in iterated form I = integraldisplay 1 parenleftBig integraldisplay 2 1 (2 x + 3 x 2 y ) dy parenrightBig dx . Now integraldisplay 2 1 (2 x + 3 x 2 y ) dy = bracketleftBig 2 xy + 3 2 x 2 y 2 bracketrightBig 2 1 = 2 x + 9 2 x 2 . But then I = integraldisplay 1 (2 x + 9 2 x 2 ) dx = bracketleftBig x 2 + 3 2 x 3 bracketrightBig 1 . Consequently, I = 5 2 . keywords: definite integral, iterated integral, polynomial function, 002 10.0 points Evaluate the iterated integral I = integraldisplay 2 1 braceleftBig integraldisplay 2 1 parenleftBig x y + y x parenrightBig dy bracerightBig dx . 1. I = 3 2 ln2 2. I = 3 ln 3 2 3. I = 2 ln3 4. I = 3 ln2 correct 5. I = 3 2 ln3 6. I = 2 ln 3 2 Explanation: Integrating with respect to y keeping x fixed, we see that integraldisplay 2 1 parenleftbigg x y + y x parenrightbigg dy = bracketleftbigg x ln y + y 2 2 x bracketrightbigg 2 1 = (ln2) x + 3 2 parenleftbigg 1 x parenrightbigg . Thus I = integraldisplay 2 1 bracketleftbigg (ln2) x + 3 2 parenleftbigg 1 x parenrightbiggbracketrightbigg dx = bracketleftbiggparenleftbigg x 2 2 parenrightbigg ln2 + 3 2 ln x bracketrightbigg 2 1 . Consequently, I = 3 ln2 . 003 10.0 points Determine the value of the double integral I = integraldisplay integraldisplay A 3 xy 2 4 + x 2 dA hyun (hh7953) – HW10 – gogolev – (57440) 2 over the rectangle A = braceleftBig ( x, y ) : 0 ≤ x ≤ 3 ,- 4 ≤ y ≤ 4 bracerightBig , integrating first with respect to y . 1. I = 64 ln parenleftBig 13 8 parenrightBig 2. I = 64 ln parenleftBig 4 13 parenrightBig 3. I = 32 ln parenleftBig 4 13 parenrightBig 4. I = 32 ln parenleftBig 13 4 parenrightBig 5. I = 32 ln parenleftBig 13 8 parenrightBig 6. I = 64 ln parenleftBig 13 4 parenrightBig correct Explanation: The double integral over the rectangle A can be represented as the iterated integral I = integraldisplay 3 parenleftbiggintegraldisplay 4- 4 3 xy 2 4 + x 2 dy parenrightbigg dx , integrating first with respect to y . Now after integration with respect to y with x fixed, we see that integraldisplay 4- 4 3 xy 2 4 + x 2 dy = bracketleftBig xy 3 4 + x 2 bracketrightBig 4- 4 = 128 x 4 + x 2 . But integraldisplay 3 128 x 4 + x 2 dx = bracketleftBig 64 ln(4 + x 2 ) bracketrightBig 3 . Consequently, I = 64 ln parenleftBig 13 4 parenrightBig . 004 10.0 points Calculate the value of the double integral I = integraldisplay integraldisplay A 2 x sin( x + y ) dxdy when A is the rectangle braceleftBig ( x, y ) : 0 ≤ x ≤ π 4 , ≤ y ≤ π 4 bracerightBig ....
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